Page 1 of 10
YOUR NAME
Department of Electrical Engineering and Computer Science
Massachusetts Institute of Technology
6.012 Electronic Devices and Circuits
Exam No. 1
Wednesday, October 7, 2009
7:30 to 9:30 pm
Notes:
1. An effort has been made to make the various parts of these problems independent of
each other so if you have difficulty with one item go on, and come back later.
2. Some questions ask for an explanation of your answer. No credit will be given for
answers lacking this explanation.
3. Unless otherwise indicated, you should assume room temperature and that kT/q is
0.025 V. You should also approximate [(kT/q) ln 10] as 0.06 V.
4. Closed book; one sheet (2 pages) of notes permitted. Formula sheet provided.
5. All of your answers and any relevant work must appear on these pages.
additional paper you hand in will not be graded.
6. Make reasonable approximations and assumptions.
assumptions and approximations you do make.
Any
State and justify any such
7. Be careful to include the correct units with your answers when appropriate.
8. Be certain that you have all ten (10) pages of this exam booklet and the three (3) page
formula sheet, and make certain that you write your name at the top of this page in
the space provided.
6.012 Staff Use Only
PROBLEM 1
(out of a possible 34)
PROBLEM 2
(out of a possible 32)
PROBLEM 3
(out of a possible 34)
TOTAL
Page 2 of 10
Problem 1 - (34 points)
This problem contains 4 independent short problems that can be worked in any order.
a) [6 pts] You find a sample of silicon, but the accompanying data sheet has been
partially destroyed so you only know that it is n-type, its resistivity is 100 Ohm-cm,
and its electron mobility is 1600 cm2/V-s. You also remember that ni(Si) = 1010 cm-3.
i) What is the approximate equilibrium electron concentration, no, in this sample?
cm-3
no =
ii) What is the approximate equilibrium hole concentration, po, in this sample?
cm-3
po =
b) [10 pts] The excess hole pop
ulation illustrated on the right
is established in an n-type
silicon sample, ND = 1017 cm-3.
The minority carrier lifetime in
this sample is 10-4 s, the hole
mobility is 640 cm2/V-s, and
the cross-section of the sample
is 10-4 cm2.
i) What is the hole current,
ih = A Jh, in this sample?
p’(x) [cm-3]
14
5 x 10
x [!m]
0
1
2
Note: 1 µm = 10-4 cm
ih =
ii) What is the total number of excess holes in this sample?
Excess holes =
Problem 1 continues on the next page
Amps
Page 3 of 10
Problem 1 continued
iii) What is the total rate at which excess holes are recombining with excess electrons
in this sample, and what is the corresponding hole current, ih,recomb?
Total recombination occurring in sample =
Hole recombination current, ih,recomb =
holes/s
Amps
c) [9 pts] Consider two p+-n diodes which are identical except for the fact that in Diode A
the minority carrier lifetime is infinite making Lh >> wN, while in Diode B the
minority carrier lifetime is finite and small enough that Lh < wN. Note: wN is the
width of the n-type side, and Lh is the minority carrier diffusion length.
i) Which of these two diodes has the larger saturation current, IS?
answer.
Diode A
Diode B
Explain your
They are similar
Explanation:
ii) Which of these two diodes has the larger total number of excess holes in the nside quasi-neutral region when a forward biased VAB is applied?
Diode A
Diode B
They are similar
Explanation:
iii) Which of these two diodes has the wider space charge layer (depletion region) in
thermal equilibrium?
Diode A
Diode B
They are similar
Explanation:
Problem 1 continues on the next page
Page 4 of 10
Problem 1 continued
d) [9 pts] This question concerns the npn bipolar transistor shown below. There is
negligible minority carrier recombination throughout except at the ohmic contacts.
Ohmic
Contact
E
Emitter
Base
n-Si, NDE =
p-Si, N AB =
5 x10 17 cm-3
1 x10 17 cm-3
Ohmic
Contact
Collector
C
n-Si
NDC = 5 x10 16 cm-3
B
i) On the axes provided, sketch and label the excess minority carrier profiles when
both junctions are forward biased with vBE = vBC = 0.6 V. Notice that this is not a
bias in the forward active region.
n’, p’ [cm -3]
x [!m]
-1
0
1
2
3
ii) Calculate the forward current gain, βf (≈ 1/δE), for this device when it is biased in
the forward active region, i.e. vBE > 0 and vCE ≤ 0. De = 40 cm2/s, Dh = 15 cm2/s.
βf =
iii) Redesign this transistor to increase the forward current gain, βf, to 100 by
increasing the doping level of one of the three regions in this device. Indicate
which region should be changed and to what the new doping level should be.
Region:
N=
End of Problem 1
Page 5 of 10
Problem 2 - (32 points)
Consider the silicon diode pictured below. It is 4 µm long, with ohmic contacts at
each end, and it is uniform p-type with NA = 1 x 1017 cm-3 for 1 µm on its far left end and
uniform n-type with ND = 1 x 1017 cm-3 on its far right end. In between these two
uniformly doped regions, the net concentration, Nd(x) – Na(x), slowly grades linearly
over a distance of 2 µm from -1 x 1017 cm-3 on the left to 1 x 1017 cm-3 on the right, as
shown in the lower figure.
Ohmic
Contact
Ohmic
Contact
A
p-Si
linearly
graded
p-Si, NA =
1x10 17 cm-3
n-Si
linearly
graded
B
n-Si, ND =
1x10 17 cm-3
x [!m]
-2
-1
0
1
2
N d(x)-Na(x) [cm-3]
1 x 1017
-2
-1
0
1
2
x [!m]
-1 x 1017
Note: 1 µm = 10-4 cm
a) [6 pts] In thermal equilibrium, what is the electrostatic potential, φ(x), in the lefthand quasi-neutral region at x = -1.5 µm, and what is the electrostatic potential, φ(x),
in the right-hand quasi-neutral region at x = +1.5 µm, and what is the built-in
potential step, Δφb, seen transiting from x = -1.5 µm to x = +1.5 µm? Use the 60 mV
rule, and log 2 = 0.3.
φ(-1.5 µm) =
V
φ(1.5 µm) =
V
Δφb =
Problem 2 continues on the next page
V
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Problem 2 continued
b) [4 pts] For the rest of this problem, a bias voltage, VAB, is applied to this diode
resulting in a total depletion region width of 1 µm, and xN =|xP| = 0.5 µm. On the
axes provided below, plot and label the net charge density, ρ(x), for x in the range - 2
µm < x < 2 µm with this bias voltage applied. Use the depletion approximation and
assume that the regions outside the depletion region are quasi-neutral, i.e. ρ(x) ≈ 0.
!(x) [coul/cm3]
-2
-1
0
x [!m]
1
2
c) [4 pts] On the axes provided below, plot and label the electric field, E(x), for x in the
range - 2 µm < x < 2 µm with the bias voltage VAB applied.
E(x) [V/cm]
d) [4 pts] What is the change in potential, Δφ, transiting the depletion region when the
bias is the same as in Part b, i.e. what is φ(0.5 µm) – φ(-0.5 µm)?
ΔφDepl.Reg. = φ(0.5 µm) – φ(-0.5 µm) =
Problem 2 continues on the next page
Volts
Page 7 of 10
Problem 2 continued
e) [4 pts] What is the change in potential, Δφ, in transiting the quasi-neutral n-type
graded region between x = 0.5 µm and x = 1.0 µm, i.e. what is φ(1.0 µm) – φ(0.5 µm)?
ΔφnQNR. = φ(1.0 µm) – φ(0.5 µm) =
Volts
f) [4 pts] On the axes provided below, plot and label the electrostatic potential, φ(x),
for x in the range -2 µm < x < 2 µm with the same bias voltage applied as in Part c.
Use the depletion approximation, and assume that the regions outside the depletion
region are quasi-neutral. In your plot make φ(0) = 0 Volts.
!(x) [Volts]
-2
-1
0
x [!m]
1
2
g) [2 pts] What is the applied bias voltage, VAB?
VAB =
End of Problem 2
Volts
Page 8 of 10
Problem 3 (34 points)
A p-type semiconductor sample with acceptor concentration NA and length L,
illustrated below, has ohmic contacts at both its ends. A light source generates MA
electron-hole pairs/cm2-s in the plane at x = XA, i.e. gL(x) = MAδ(XA). Assume low-level
injection and quasi-neutrality everywhere in the bar.
gL(x) = MA !(XA)
Ohmic
Contact
Ohmic
Contact
p-type
NA
x
0
XA
L
The general equation governing the excess minority carriers in a uniformly doped
material is
d 2 n'(x)
n'(x)
1
"
= "
gL (x)
2
2
dx
Le
De
a) [4 pts] What boundary condition is imposed on the excess minority carriers n’ at x =
0 and x = L ?
!
Boundary condition at x = 0:
Boundary condition at x = L:
b) [4 pts] We now make the assumption that the minority carrier lifetime is very long,
which simplifies the general equation to:
d 2 n'(x)
1
" #
gL (x)
2
dx
De
What quantitative restriction is placed on the minority carrier lifetime, τe, for this
assumption to be valid?
!
τe >> ≈ <<
(circle one)
Problem 3 continues on the next page
(fill in the blank)
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Problem 3 continued
c) [6 pts] Using the long-lifetime approximation in part (b), determine two constraints
(i.e. boundary conditions) on the excess minority carriers at x = XA, i.e. relating
n’(XA− ) to n’(XA+).
Constraint on n’ at x = XA:
(Hint: Finite currents imply finite dn’/dx.)
Constraint on dn’/dx in the vicinity of x = XA:
(Hint: What goes in must come out.)
d) [6 pts] On the axes provided sketch the excess minority carrier concentration, n’(x),
everywhere inside the semiconductor. Label your sketch with the relevant
equations for n’(x).
n’(x)
x
0
XA
L
e) [6 pts] On the axes provided sketch the minority carrier diffusion current, Je,diff(x),
everywhere inside the semiconductor. Label your sketch with the relevant
equations for Je,diff(x).
J e,diff (x)
0
XA
Problem 3 continues on the next page
x
L
Page 10 of 10
Problem 3 continued
f) [4 pts] In the space below, briefly explain (approx. 25 words or less) why the
minority carrier diffusion is the dominant minority carrier current.
g) [4 pts] A second light source is added illuminating a single spot along the
semiconductor at x = XB, where xB > XA, and generating electron-hole pairs at a rate
MB, so that gL(x) is now
gL (x) = M A " (X A ) + M B " (X B ) .
Find n’(x) and Je,diff(x) everywhere inside the semiconductor under this new
illumination condition. If you could not do Parts d and e, indicate how you would
use the results of!those parts to answer this question.
n’(x) =
Je,diff(x) =
End of Problem 3
End of Exam One
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6.012 Microelectronic Devices and Circuits
Fall 2009
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